The radii of curvature of both surfaces of a convex lens of focal length $f$ and focal power $P$ are equal. One of the surfaces is made plane by grinding. The new focal length and focal power of the lens are respectively:

The radius of curvature of each surface of a convex lens having refractive index $1.8$ is $20 \ cm$. The lens is now immersed in a liquid of refractive index $1.5$. The ratio of power of the lens in air to its power in the liquid will be $x : 1$. The value of $x$ is $.....$

In an optical bench experiment, the index error is $+1 \, cm$ and $-1 \, cm$ between the object needle and lens, and between the lens and image needle, respectively. The observed values of $u$ and $v$ are $9 \, cm$ and $17 \, cm$. The focal length of the lens is ..... $cm$.

What is the focal length (in $cm$) of a biconvex lens with a radius of curvature of $40\, cm$ and a refractive index of $1.65$?

$A$ bi-convex lens has a radius of curvature for both surfaces equal to $1/6 \ cm$. If this lens is to be replaced by another convex lens having different radii of curvature on both sides $(R_1 \neq R_2)$,without any change in the lens power,then the possible combination of $R_1$ and $R_2$ is:

Radii of curvature of a converging lens are in the ratio $1 : 2$. Its focal length is $6 \,cm$ and refractive index is $1.5$. Then its radii of curvature are ......... respectively.